amys excerpt

Advanced Math for Young Students

A First Course in Algebra

By Philip Keller

Copyright Philip Keller, 2014

All rights reserved

CONTENTS

Introduction 5

UNIT 1: Charts, Graphs and Rules 9

Chapter 1: The Mathematical Match-Makers 10

Chapter 2: Functions and their Graphs 29

Chapter 3: “Rules, man. What ARE rules?” 47

UNIT 2: The Search For 𝒙 Begins 71

Chapter 4: “I’m thinking of a number.” 72

Chapter 5: “Now I’m thinking of TWO numbers!” 103

UNIT 3: Relationship Advice 141

Chapter 6: Proportions 142

Chapter 7: The Ups and Downs of Inverse Variation 169

Chapter 8: Exponential Functions and their Inverses 183

Afterword: One More Unusual Proportion 221

Answers and Selected Hints 225

Index 257

Introduction 5

INTRODUCTION

I wrote this book with several different groups of students in mind.

For 26 years, I have been a high school physics teacher. I work in an excellent, well-

regarded high school and I have been fortunate to have many talented students who

soak up all the physics I can teach them, and more. But every year, I also teach

students who struggle to master the topic, despite their great efforts and mine. And

I know from discussions with colleagues, both within my school and across the

country, that we are not the only ones struggling. There is something getting in our

way. Maybe this will seem obvious to anyone who has struggled in physics, but

here’s what I think: I think it’s the math.

Physics applies math. It’s all about finding relationships and solving the puzzles

that the laws of physics present. For the most part, this work is done in the

language of mathematics, and more specifically, the language of algebra. So to be

comfortable learning physics, a student has to be fluent in that language. Algebra

cannot just be a memorized set of procedures for finding ‘x’. It has to be a symbolic

way of representing ideas. But for many students, that level of fluency is not

attained in just one year of algebra – which is all that many students have had

when they start studying physics. It’s no wonder that some struggle.

It is not only physics students who struggle. For even more than 26 years, I have

been teaching students how to prepare for the math portion of the SAT. What I

have seen over the years is that most students are not fluent enough in algebra to

successfully apply algebra on the SAT. One goal of my SAT course is to teach

alternative, non-algebraic approaches to SAT problems. It is also a major theme of

my math SAT book, The New Math SAT Game Plan. And I will tell you something

you may find surprising (or even distressing): on the SAT, these non-algebraic

methods work very nicely. They won’t get you to an 800, but they will take you

pretty far. And even my top scorers report that they like to mix in the non-algebraic

methods along with the standard approaches (which, as top students, they also

know how to use).

The non-algebraic methods, however, won’t get you very far in physics. In fact, a

student who does not really learn the language of algebra is going to struggle in all

later math and science classes: physics, statistics, computer science and beyond.

That STEM door is swinging closed because one year of algebra class did not lead to

sufficient fluency. So why spend only one year? Why not start earlier?

6 Advanced Math for Young Students

I am not saying that every 7th grader should be in a high-school version of Algebra I.

But I am saying that every middle school student should, over the course of the

middle school years, start learning about and thinking about the ideas of algebra

(even some ideas that won’t reappear until Algebra II or Pre-calculus). These are

ideas that take time to ponder. Even during that one year of Algebra I, so much

time has to be spent learning the mechanical side of algebra that there is barely

time to think about the “why” and “what for.” I wrote this book to solve that

problem. Algebra is more interesting and easier to learn when you understand what

you are doing and why you are doing it.

There is one other group of students I have been thinking about while writing this

book. These students are the puzzle-solvers, the thinkers, the ones who ask the

challenging or off-beat questions. They are in middle school or even younger and

they are hungry to learn math beyond arithmetic. In fact, these students are not

always that good at arithmetic. But they do love to think. I hope this book gives

them plenty to think about.

What you will find in this book:

Unit 1 introduces the idea of a mathematical function. We will talk about

combining functions, composition of functions and inverse functions. We will see

how functions can be represented by graphs. Also, we use this setting to introduce

the key abstraction of algebra: the representation of numbers with letters. We will

also introduce algebraic expressions in this context.

Unit 2 focuses on using algebra to “find the mystery number.” So we will introduce

algebraic equations and what it means to solve them. We’ll also apply these tools to

solve the dreaded “word problems.” Then, we will look at linear equations with two

variables and the ordered pairs that comprise their solutions. This will lead us back

to functions again. We’ll also examine systems of linear equations graphically and

in word problems as well.

Unit 3 explores some of the ways that the language of algebra is used to express

mathematical relationships. In this unit, you can tell that I am trying to prepare

my future science students. We’ll start with direct proportion, where we will also

learn to solve the classic “ratio problems.” Then, we will extend those ideas to

“direct square” variation. Next, we will explore inverse variation, and then inverse

square variation. For all of these, we will look at the graphs, the equations and the

ratios involved. I will include applications but no specific physics knowledge will be

Introduction 7

pre-supposed. (Some physics knowledge might be imposed, but I will try to keep

that to a gentle minimum.) The closing chapter is devoted to exponential functions.

This will lead us to an introduction to logarithms, because when I said advanced

math for young students, I really meant it.

What you WON’T find here:

This is most definitely not an arithmetic book. I assume that you know what

decimals and fractions are and that you can do some basic arithmetic with them, but

if you can’t, that doesn’t have to hold you back. Use a calculator when you can’t do

the arithmetic by hand.

This is not the last algebra book you will ever need. You will learn a lot of algebra

here, and I hope you find the discussions to be slow and thorough so that you

actually understand what we are doing. But there are topics that are traditionally

part of Algebra I that we won’t be discussing. For example, we will be working

mostly with linear equations – no quadratic formula to be found here. Also, no

factoring or “foiling.” I have to leave some of the fun for your high school years.

And, speaking of fun during high school…

This book is dedicated the group of outstanding math teachers I was lucky to have

at Hunter College High School from 1976 to 1982: Mr. Nadell, Dr. Bumby, Mr.

Klutch, Ms. Barnhart, Ms. Krilov and Dr. Ruderman. Thank you all for the

enthusiasm, clarity and friendly sense of humor that you each brought to the

classroom. And thank you also for the innovative curriculum. In the math classes I

took in college and even in graduate school, I saw material that I had already

learned back in “Experimental Math” at Hunter.

I also want to thank my wife, Daphne, and my children, Reuben and Jane. In

addition to their familial support, they generously contributed their time, their

suggestions and their editing skills. In addition, my thanks to Dr. Yvonne Lai,

currently of the University of Nebraska and formerly a student at Holmdel High

School, for her many useful insights and interesting ideas for exercises. And thank

you also to Kemy Lin of Princeton University’s Student Design Agency for her cover

design.

Philip Keller

[email protected]

9

UNIT 1: Charts, Graphs and Rules – An Introduction to Functions

We will begin our study of algebra by studying something

called a “function.” In everyday language, the word “function”

refers to what an object is used for, its purpose, or how it

works. For example, the function of a can-opener is (no big

surprise here) to open cans. The function of a lawn-mower is

to cut grass. That’s what it does when it’s “functioning.” You

can certainly make your own list of useful objects and their

functions.

But in mathematics, the word “function” has an entirely different meaning. We use

it to describe a type of match-making process that is central to much of

mathematics. It would be difficult to over-state how important this concept is going

to be as you continue to study math and its applications. So that’s going to be one of

the major goals of this book: to teach you to think about functions the way a

mathematician does.


Chapter 1: The Mathematical Match-Makers

1.1 Making assignments

The best way for me to explain what a function is is to tell you exactly what a

function does.

A function takes the members of one group and assigns each member of that first

group to a member of another group. In this way, it is very much like a match-

making machine. The groups are most often sets of numbers, but they don’t have to

be. You can learn a lot about functions by exploring ways of assigning members of

one group to those of another, working with things that aren’t numbers.

So for our first function, let’s look at an example involving days and hats. Suppose

you decide to wear a hat to school every day according to the following schedule:

You can call this match-making scheme a “schedule”,

but mathematically, it also meets the requirements to

be what we call a “function.” It takes the members of

one group (the days of the week) and assigns each one

to a member of the other group (the types of hats).

That is all it takes to be a function.

We have special names for these two groups that are involved in any function. The

first group, whose members will be assigned, is called the “domain” of the function.

The second group, to which the domain is assigned, is called the “range.”

DAY HAT

TYPE

Monday Baseball

Tuesday Fedora

Wednesday Beret

Thursday Cowboy

Friday Ski Cap

The Mathematical Match-Makers 11

So in this example, the “domain” is just the days of the week that are being assigned

hats:

Monday, Tuesday, Wednesday, Thursday, and Friday *.

And the “range” is the collection of hats that the days are being assigned to:

Baseball, Fedora, Beret, Cowboy, and Ski Cap.

For now, we just want to be able to answer basic questions about this function such

as:

What hat type does this function assign to the day “Wednesday”?

What day is assigned the “Ski Cap”?

Another way of talking about this is to think of the function as a machine. When we

INPUT a day of the week, the function OUTPUTS a hat type. It’s as if you just

couldn’t keep track of what hat to wear each day, so you bought an expensive

machine to do it for you. You write down a day of the week on a piece of paper.

Then you put the paper INTO the machine. So we say that the day is the INPUT.

Then, the machine chugs away for a while, doing whatever internal magic it has to

do before finally spitting out a new piece of paper with a type of hat printed on the

paper. So the hat type is the OUTPUT of the machine. (Of course, for far less

money you could just make a chart.)

So using this new machine vocabulary, the answers to the two questions above are:

When the INPUT is “Wednesday”, the OUTPUT is “Beret.”

When the OUTPUT is “Ski Cap”, the INPUT must have been “Friday.”

* The domain and range are both examples of “sets.” There is a more formal notation for

writing sets that you will see in the next section.


Now, your friend may also want to make his own Hat Function. But suppose he has

a less extensive hat collection. His function might look like this:

Notice that the domain is the same as the previous

function:

Monday, Tuesday, Wednesday, Thursday, and Friday.

But the range is not as big. It’s just three hats.

Baseball, Fedora, and Cowboy.

But does this schedule even qualify as a function? The answer is that yes, it does.

Each member of the first set (the domain) is assigned to a member of the second set

(the range). You may not like it that more than one member of the domain is being

assigned to the same member of the range, but that does not break any math rules.

We never required that each member of the domain had to have a member of the

range to call its own. Your friend’s hat function assigns multiple members of the

domain to the SAME member of the range. We are going to allow that kind of

multiple assigning and still call this a function.

(It may seem arbitrary to you that we allow this. More on this later…)

On the other hand, there is something that seems similar that we absolutely do

NOT allow functions to do. A function cannot assign one member of the domain to

more than one member of the range. As someone new to this concept, you may not

see why this isn’t allowed. After all, it’s easy to imagine a schedule like this:

It’s just like the first example except that on

Wednesday you have a choice. But while this is

allowed for schedules, it is NOT allowed for

functions. Functions make assignments – they do

not give you options. That is the main rule of

functions: every input gets assigned ONE output.

We’ll call that the “assignments and not options”

rule.

DAY HAT

Monday Baseball

Tuesday Fedora

Wednesday Baseball

Thursday Cowboy

Friday Cowboy

DAY HAT

Monday Baseball

Tuesday Fedora

Wednesday Beret or Top Hat

Thursday Cowboy

Friday Ski Cap

PAGES 13 - 28 ARE OMITTED IN THIS EXCERPT

Functions and their Graphs 29

CHAPTER 2: Functions and their Graphs

In Chapter 1, you saw that functions assign each member of their domain to a

member of their range. We showed how a given function made those assignments

by presenting all of the information in a chart, using one column for the “inputs”

(the members of the domain) and another column for the “outputs” (the target

members of the range). Now we are going to explore another way of displaying that

same information. This new way only works when the domain and range are

numbers – but from this point on, all of our examples will in fact be numeric.

2.1 How to Construct the Graph of a Function

The graph of a function is a visual way to represent the relationship between the

domain and the range. Instead of a chart, you get a picture that contains the same

information but makes it easier to recognize trends. Graphs also let us identify

specific types of functions on sight.

So how do we generate these pictures? Let’s go back and look at our old friend Fred.

Like every function, Fred matches inputs with outputs. That’s what we see in

Fred’s chart. Now, to turn the chart into a picture, we just have to think of each

input-output pair as an “ordered pair,” the kind you can plot on the coordinate

plane. So each input-output pair is now represented by a dot on that plane. Taken

together, the dots make up a picture.

INPUT: x OUTPUT: f(x) --->

ORDERED

PAIR ---> GRAPH

0 3

(0,3)

1 5

(1,5)

2 7

(2,7)

3 9

(3,9)

4 11

(4,11)

5 13

(5,13)

6 15

(6,15)

7 17

(7,17)

So now we have a way of visualizing the function Fred. We call it the graph of the

function. And in this case, we also say that this function is “linear.” You can look at

the graph and see why we say that.

0

2

4

6

8

10

12

14

16

18

0 1 2 3 4 5 6 7 8


This way of looking at functions is a BIG IDEA in mathematics. It is worth

reviewing what we just did. (Go through this list slowly to make sure you

understand each item.)

Functions assign “inputs” to “outputs.”

Each input and its associated output can be matched together to make an ordered

pair: (input, output)

Each ordered pair can then be graphed on the coordinate plane. So each pair

makes a dot on the graph.

The collection of dots is referred to collectively as “the graph of the function.”

Each point on the graph of a function tells you how one particular input (the x-value

of that point) is assigned its output (which is the y-value of that same point).

DOT, DOT, DOT…LINE?

The example we just did leads us immediately to a deep philosophical debate:

Should we connect the dots? The short answer is: Well, usually we will, but not

always.

Suppose we have been given a chart for some numerical function. And suppose we

have already interpreted each input–output pair as an ordered pair and we have

already plotted each point on the graph. Now…


Rules, man. What ARE rules? 37

Look again at the question:

What is h(2)?

2 is the input to the function.

The input is the x-coordinate of the ordered

pair. So start on the x-axis, at 𝑥 = 2.

Trace vertically along the 𝑥 = 2 line until you

meet the graph.

Trace horizontally to the y-axis to see what the

output was. This time, it was 5. So ℎ(2) = 5.

We started on the x-axis because we had been given the INPUT value.

But now let’s look again at this question: If 𝒉(𝒂) = 𝟐, what is the value of 𝒂?

This is not the same question as the last one!

We have been given different information this

time. We have been told the OUTPUT of the

function.

So 2 is the y-coordinate of the ordered pair.

That’s why we start on the y-axis at 𝑦 = 2.

Trace along HORIZONTALLY until you meet

the graph of the function.

Trace vertically down to the x-axis to see what

the input must have been.

This time, it was 8. And since ℎ(8) = 2 , 𝑎 = 8 is the value we were looking for.

0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8

0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8


KEY IDEAS IN THIS SECTION

Once you know how to read the graph of the function, you can solve the same kinds

of puzzles you solved with a chart.

In particular, given an input, you can find the output. And given the output, you

can find the input. (But sometimes, you have to estimate as best as you can from

the graph.)

EXERCISES

Use the graph of the function g, shown at right,

to answer exercises 13 – 17. Note the scale on

the x-axis.

13. Find each of these values:

a. 𝑔(−4) b. 𝑔(0) c. 𝑔(4)

14. If 𝑔(#) = 3, then what value does the

symbol # represent?

15. What is the value of 𝑏 if 𝑔(𝑏) = 𝑏?

16. What are the coordinates of the point on

the graph that represents an input of zero?

17. What are the coordinates of the point on

the graph that represents an output of zero?

Rules, man. What ARE rules? 39

Exercises 18 - 27 are based on the graph of

function f, shown at right. The domain is

all of the values from -4 to 10, not just the

points that have been emphasized with

dots.

Find each of the following values:

18. 𝑓(0) 19. 𝑓(6)

20. 𝑓(−2) 21. 𝑓(10)

22. If 𝑓(𝑚) = 12, find 𝑚.

23. If 𝑓(𝑘) = 3, and 𝑘 < 6 find 𝑘.

24. If w represents an integer and 𝑓(𝑤) = 7, list all possible values of 𝑤.

[Hint – don’t forget to look both ways…]

25. What is the smallest integer, 𝑎, for which 𝑓(𝑎) < 𝑎 ?

26. If 𝑏 is an integer and 𝑓(𝑏) = 𝑓(1), find all possible values of 𝑏.

27. Does this function have any negative inputs? Where would you look to find

them? Does it have any negative outputs? Where would you look to find them, as

well?

28. If 𝑧 is an integer and 𝑓(𝑧 + 1) = 𝑓(𝑧) + 1, find all possible values of 𝑧.

29. Draw the graph of a function, 𝑔 for which it is the case that 𝑔(𝑥) > 𝑓(𝑥) for all 𝑥

in the domain.

30. Looking at the graph of 𝑓, would you expect this function to be invertible? (This

will be discussed further in the next section.)



Chapter 3: Rules, man. What ARE rules?

So far, we have seen that functions assign inputs to outputs. We can represent

those assignments in a two-column chart or on a two-axis graph. The chart and the

graph are presenting the same information but in different formats. You may

already have been wondering how the function knows what numbers to assign. As

you may have suspected, there often is a simple procedure that a given function uses

to assign the matches.

3.1 Using Rules to Assign Inputs to Outputs

We will start with a really basic example and we are going to state the procedure in

the form of a sentence:

Let the function f assign to each of its inputs an output value that is one greater

than the input.

This rule is useful. We can use it instead of a chart or graph. If someone asks us:

What is the value of 𝒇(𝟖)?

We can just follow the given rule: 8 is the input. The output will be one more than

that, or 8 + 1 = 9. Thus, 𝑓(8) = 9.

In fact, if we prefer charts, we can easily generate a chart for the function, f. We

just have to pick a handful of values to use as the inputs and then follow the rule of

this function to get the outputs.

Then, we can make ordered pairs and

create our graph:

INPUT: x OUTPUT: f(x)

0 1

1 2

2 3

3 4

4 5

5 6

6 7

7 8

8 9

0

1

2

3

4

5

6

7

8

9

10

0 1 2 3 4 5 6 7 8

41

For example, when the input is 3, the rule assigns the output 3 + 1 = 4, just as you

see highlighted in the chart. Then, that makes the ordered pair (3, 4) which is one

of the points on the graph of the function. I have circled it for emphasis.

So now we have three ways to represent the function 𝑓: chart, graph, and rule.

Mathematicians use all three of these all of the time. But they don’t often write the

rules out in a sentence the way we did. Let’s take another look at that rule and see

if we can’t be more concise.

USING ALGEBRAIC EXPRESSIONS TO STATE THE RULES

Instead of saying:

“Let the function f assign to each of its inputs an output value that is one greater

than the input,”

we are going to say:

“Let the function f be defined by 𝒇(𝒙) = 𝒙 + 𝟏."

That sentence has a lot of meaning packed into it. You are going to see sentences

like it many times in your mathematical future. Let’s take a closer look…

When it says…. That actually means…

“Let the function 𝑓…” We are getting ready to introduce a function.

Its name is “𝑓.”

“…be defined by…” Now we are about to give you the rule that

this function uses to assign each input to its

output.

.”..𝑓(𝑥) = …" When we tell you the rule, we are going to

use the letter '𝑥' to represent the number

that is input into the function.

"…𝑥 + 1" Finally! At the end of the sentence, we tell

you just exactly what to do with the input.

In this case, we just want you to add 1 to it

and that will give you the value of the

assigned output.


This sentence is such a common and formal sentence that is almost spell-like. Say

the words properly and you conjure into existence a mathematical function! But you

have to say it right. It may help you to understand this better if you follow along

with me as I walk through the process.

CONJURING A NEW FUNCTION

I’m thinking of defining a new function. I want it to assign each input

to the number you get when you multiply the input by itself. How can I

say this concisely? Well, first I should pick a letter to use as the name

of this function. I pick ‘g’. So my sentence will begin: “Let the function

g be defined by…”

Now I have to pick a letter to represent the input to the function. I can pick any

letter I want but the letters used most often for this purpose are ‘x’ and ‘t’. It is

especially helpful to beginners if we use standard letters for certain purposes. (But

we don’t want that to become a crutch, so later in the exercises, I’ll be sure to mix

things up.) For now, as my input letter, I’ll use the letter ‘t’. Now my sentence

reads: “Let the function g be defined by g(t) = …”

All I need now is the mathematical expression that means “multiply the number by

itself.” That’s just raising to the second power (or “squaring”). So I can finally finish

that sentence:

Let the function g be defined by g(t) = t2.

When you understand that sentence, you automatically think: “OK, the function is

named g. And it takes inputs and squares them to get the outputs.” And now, it’s

easy to make a chart and a graph for this new function. When you do…

INPUT: x

OUTPUT:

g(x) --->

ORDERED

PAIR

-3 9

(-3,9)

-2 4

(-2,4)

-1 1

(-1,1)

0 0

(0,0)

1 1

(1,1)

2 4

(2,4)

3 9

(3,9)

-4 -3 -2 -1 0 1 2 3 4


71

UNIT 2: The Search for 𝒙 Begins

Writing and Solving Equations

There are times in math when you know something about a

number but not the value of the number itself. We saw this

several times in our study of functions. It was like playing a

game: find the mystery number. Well, a big part of

learning algebra is developing methods for playing exactly

that game. It is traditional to use the letter ‘x’ to represent

the mystery number. Many people will tell you that all they remember from their

time in high school algebra is the endless struggle to find 𝑥.

You can use these methods to find the input when we know the output of a function.

But these methods have many other applications as well. They can be used to solve

the kinds of puzzles that you encounter in physics, chemistry, economics and (of

course) higher level mathematics. So before we return to the problem of finding the

input to a function, we will practice using these methods on a variety of what are

universally (but oddly) called “word problems.”

One other note before we jump in. When you see the examples, you may be tempted

to figure out the mystery number by other methods. If you are resourceful, you can

often solve these puzzles without “doing algebra.” In general, I encourage that kind

of evasion*. But in this case, remember that the point of this unit is to learn the new

skills. There are plenty of puzzles in the world of math and science that are too

hard to solve any other way. Besides, when you have mastered this method, you

will see that it is a quick and powerful way to cut through the mystery.

* And when I am teaching SAT math, I encourage it a lot.


Chapter 4: “I’m thinking of a number.”

4.1 Equations and Solutions

We begin with a puzzle.

I’m thinking of a number. If you triple the number that I am thinking of and then

reduce the resulting product by 7, you get a number that is 3 more than what you

started with.

This is a perfectly fine English sentence, but we are going to translate it into the

language of algebra. To do this, we pick a letter to represent the number that I am

thinking of. And we’ll go along with tradition and call that number ‘x’. Then, we

have to replace words with expressions.

Where it says, “If you triple the number that I am thinking of and then reduce the

resulting product by 7” we can write: 3𝑥 – 7.

Where it says, “a number that is 3 more than what you started with”, we write:

𝑥 + 3.

Those two phrases are connected with the phrase, “you get a number that is”, which

can be represented with a single character: the “=” sign.

Putting the pieces together, we write 𝟑𝒙– 𝟕 = 𝒙 + 𝟑, which is a much more concise

way of expressing the exact same puzzle we started with.

It is also an example of an algebraic equation. It is an “equation” because it is a

statement that two expressions are equal. And it is “algebraic” because at least one

of the expressions (and in this case, both of them) contains an algebraic “variable.”

The variable, in this case, is the letter “x”, which is representing a number that we

have yet to determine. And the equation 𝟑𝒙 – 𝟕 = 𝒙 + 𝟑 is the clue that we use to

find that mystery number.

The equation says the same thing as the “word problem” but in a concise, symbolic

way. Though you may prefer to see it written out in words, there is a major

advantage to writing this sentence as an equation (aside from its being shorter).

You can do things with the symbols that you can’t do with the words, some very

useful things.

When you have an equation, there are ways that you are allowed to manipulate the

equation to make it simpler. This changes the equation so that it looks different –

“I’m thinking of a number.” 73

in fact it IS different. But as long as you follow the rules about what trans-

formations are allowed, each new equation that you get will be “equivalent” to the

one that you started with.

For example, one of those permitted transformations will take our starting equation

and change it so that instead of 3𝑥 – 7 = 𝑥 + 3 we now have:

𝟑𝒙 = 𝒙 + 𝟏𝟎.

Please don’t worry (yet) about what we did to change the first equation into this new

one*. Right now, I just want to convince you that the two equations are “equivalent”

in the sense that they have the same “solution.”

WHAT IS A “SOLUTION”?

A solution of an algebraic equation is a number that you can use to replace the

variable, resulting in a true mathematical sentence. We need to find a number that

makes

𝟑𝒙 – 𝟕 = 𝒙 + 𝟑 into a true statement.

The process of finding that solution is called “solving the equation.” But how do you

actually do this? Essentially, you continue manipulating the equation in the

allowable ways until you find a way to transform the original equation to the form:

x = a number

at which point, you have solved the equation. The variable is “isolated” by itself on

one side of the equation and there is a number on the other side. The number is the

solution.

The trick is to learn how to get from the original equation

to the final one. In some ways, it’s like one of those word

ladder puzzles where you are trying to transform one word

into another by changing one letter at a time. In that

game, there is only one rule: every change has to produce

a real word. And in this game, there is also only one rule:

every change has to produce an equation that is

equivalent to (has the same solution as) the original one.

* In fact, what I did was add 7 to both sides. As for why that is allowed, stay tuned. And

don’t panic.

Here is a word ladder

that changes HATE to

LOVE:

HATE

LATE

LANE

LONE

LOVE


One “Equation Ladder” for this equation looks something like this*:

3𝑥 – 7 = 𝑥 + 3

3𝑥 = 𝑥 + 10

2𝑥 = 10

𝑥 = 5 Look! This is an equation in the form “x = a number.” That is the

solution we were seeking.

Again, let me be clear: Do not worry if you don’t know how we did that yet! You will

learn how to do it yourself very soon. The point of this example was to introduce the

basic ideas, the vocabulary, and the process that will let us solve word problems.

Let’s review that process:

We started with a “word problem” which provided (in words, of course) information

about a number whose value we had yet to determine.

We expressed the provided information in the form of an equation: 3x – 7 = x + 3

We manipulated that equation to get new equations, such as 3x = x + 10.

Eventually, we manipulated it further to get x = 5.

Once we have “variable = number”, we have solved the equation. We have found the

mystery number. And in this case, the number was 5.

Verifying Solutions

You don’t have to trust me about any of this. You can “verify” the solution. This

just means that you go back to the original equation and replace the variable with

our solution. If 𝑥 = 5 is in fact a correct solution, we’ll get a true statement.

3𝑥– 7 = 𝑥 + 3

3(5) – 7 = 5 + 3

15 – 7 = 8….and yes, that is true!

* Like word ladders, there is often more than one way to go from the start to the finish.



Some tables in the cafeteria have room for 6 students and others have room for 10

students. There are a total of 12 tables and they can hold a total of 100 students.

How many tables of each variety are there?

Our first step is to translate this question into algebra. We begin by choosing a

letter to represent one of the amounts of tables. Let’s let 𝑥 = the number of tables

that hold 6 students. Since the total number of tables is 12, if there are ‘𝑥’ of one

kind then there must be 12– 𝑥 of the other kind of table, the ones that hold 10

students. Then, to find the total number of seats, we multiply the number of each

type of table by how many students it can hold.

6𝑥 + 10(12– 𝑥) is that total. Since the problem has told us that there are seats for

100 students altogether, we can write the equation:

6𝑥 + 10(12– 𝑥) = 100

Now, we just have to solve this equation. Here is one ladder that works:

6𝑥 + 10(12 − 𝑥) = 100 This is my starting equation.

6𝑥 + 120– 10𝑥 = 100 I saw a chance to distribute so I tried it.

6𝑥 + 120 = 100 + 10𝑥 I was going to combine the terms. But I didn’t want to deal with negative numbers (though there is nothing wrong with doing it that way). I decided instead to “move things over” to the other side by adding 10x to both sides.

20 = 4𝑥 OK, I did two things in one step: I subtracted 100 from both sides and I also subtracted 6x from both sides.

5 = 𝑥 I divided both sides by 4 and I was done.

If I did this right, then there are 5 tables that hold 6 students. So there must also

be 12 – 5 = 7 tables that hold 10 students. And since 5x6= 30, 7x10 = 70 and 30 + 70

does in fact equal 100, the algebra has led us to the correct answer.

Two things to note about this example:

1. My equation ladder is not the only ladder that works. If you solved this

differently but landed on x = 5, then your ladder is also correct.

2. Even my original equation is not the only way to start. Later in this book,

you’ll see another method for solving problems like this one. The first step is

the hardest, especially when you are a beginner. If you are having trouble

seeing how to start these kinds of problems, there is a method you can use to

teach yourself what to do by starting with numbers and then switching to

variables. Let’s take a look.


LOST IN TRANSLATION?

Here is a trick to help you see how to come up with the starting equation.

Pretend you were going to solve this problem by random trial and error. You might

start by trying to guess how many 6-student tables there are. Let’s guess 3. Then,

how could you check if you were right?

Well, if there are 3 of the 6-student tables, there must also be 12 – 3 = 9 of the 10-

student tables. Now let’s see how many seats that gives us: 6×3 + 10×9 = 18 + 90

which is NOT 100.

Now, you could just keep guessing but you would be missing the point! Instead, go

back and call your original guess ‘𝑥’. Then, do to the variable ‘x’ exactly what you

did to the number you used as your first guess when you were checking to see if it

was right.

So instead of 12 – 3 = 9, you would now have the expression 12– 𝑥. And instead of

6×3 + 10×9, you would have the expression:

6𝑥 + 10(12– 𝑥).

And finally, instead of checking to see if it equals 100, you set it equal to 100, giving

you:

6𝑥 + 10(12– 𝑥) = 100

And now, you are ready to solve it with an equation ladder!

Using concrete numbers while you are first figuring out how to solve the puzzle is a

good way to get started. Then, the same steps you followed to check if your guess

was right are the steps you will follow algebraically to set up an equation.

We will start the next example by using this method to translate.


John had a collection of x marbles. He gave away 1/3 of them to his little brother.

Then he lost 20 of them somewhere in the back of the family car. Now he has ¼ of

the marbles he originally started with. How many did he start with?

First, we take a guess. I pick 60. Is it right? Well, if he gives away 1/3 of them,

that’s (1/3) of 60 which is 20. That leaves him 60 – 20 = 40. Then he loses 20 more

so now he has 40 – 20 = 20. Is that ¼ of what he started with? No, ¼ of 60 is 15, not

20.

Now, instead of more random guessing, I call my guess 𝑥. Then I do to the variable

𝑥 all of the things I did to my original guess: Find 1/3 of 𝑥. Subtract that from the

starting amount. So far, that leaves me with 𝑥 – (1/3) 𝑥. Now subtract 20 more lost

marbles. That takes us to 𝑥 – (1/3)𝑥 – 20. If our number is correct, we’ll get the

same result when we find ¼ 𝑥. So let’s say that in the form of an equation and then

solve it.

The equation is 𝑥 − 1

3𝑥 − 20 = 1

4𝑥

One possible equation ladder follows below, with step-by-step explanations.

𝑥 − 1

3𝑥 − 20 = 1

4𝑥

If you read the problem again, you will start to see how the starting equation just restates the problem, using the language of algebra.

3𝑥 − 𝑥 − 60 = 3

4𝑥

In this step, and in the next, I did something to avoid having to work with fractions. First, I multiplied both sides by 3, distributing on the left side…

12𝑥 − 4𝑥 − 240 = 3𝑥

…and then I multiplied both sides by 4, again distributing. I did this because the fractions had 3 and 4 in the denominators. I could also have done it in one step, multiplying both sides by 12. This trick is sometimes called “clearing the fractions.”

8𝑥 − 240 = 3𝑥

Here, I combined the like terms on the left-hand side.

5𝑥 = 240

Then, in one step, I subtracted 3x from both sides AND added 240 to both sides.

𝑥 = 48

Divide both sides by 5 and the ladder is complete.

Once again, we should verify that our solution is correct.

John starts with 48 marbles. He gives away 1/3 of them, which is 16 marbles,

leaving him with 48 – 16 = 32 marbles. Then he loses 20, leaving him with 12. And,

like magic, 12 is ¼ of the 48 he started with.


KEY IDEAS

You can perform addition, subtraction, multiplication or division as long as you “do

it to both sides” of the equation. (And you can’t multiply or divide by zero.) These

are transformations that will produce a new equation that is equivalent to the

original.

Word problems can be translated into algebraic equations. Then, the legal

transformations can be used to solve the equation.

EXERCISES

Solving puzzles is more interesting than just solving equations out of context. So in

these exercises, I am giving you the puzzles to solve. But if you can’t come up with

the equations yourself yet, look at the answer key in the appendix. I give the

equations first and then, separately, the suggested ladders and solutions. That way,

even if you have to peek to get started, you still get to practice your equation solving

skills. So, for each of the following:

a. Translate the puzzle into the language of algebra, looking at the appendix as

needed.

b. Solve the equation.

c. Check to see if your solution actually answers the question.

27. A number is tripled and that product is reduced by 10. The result is the same as

the original number. What was that number?

28. Half of the marbles in a jar were red. After 20 red marbles were removed, only

one quarter of the remaining marbles were red. How many marbles were in the

jar to begin with?

29. Cars and motorcycles pay $5 per wheel to ride on a ferry. If 24 vehicles paid a

total of $380, how many of those vehicles were cars?

30. The sum of 3 consecutive EVEN integers is 612. Find the smallest of the three

integers.

31. The sum of 4 consecutive ODD integers is 45 more than the largest of the four

integers. Find the largest of the four integers.


32. The price of a sweater increases by 20% one week and then by $10 the following

week. The final price is twice as big as the original price. What does the

sweater cost now?

33. In a school club, there are twice as many sophomores as freshman and twice as

many juniors as sophomores and 12 seniors. If the club has 68 members overall,

how many of them are juniors?

34. A cell phone provider offers two text-message plans. You can pay $5 per month

and then $.02 per text or $8 per month and then $.01 per text – and you can

choose which plan you prefer each month! (Real plans never work like that.) At

the end of the month, you discover that either plan costs you the same amount.

How many texts did you send that month?

35. When a train averages 60 mph for a trip, the trip takes 15 minutes (or ¼ of an

hour) less than when the train averages only 40 mph. How many miles long is

the trip?

36. A school survey reveals that the ratio of students taking Latin to students

taking Spanish to those taking French is 2:5:3. If the total student enrollment

in these three classes is 300, how many more students take Spanish than

French?

37. John is three times as old as Mary. In 10 years, he will be twice as old as Mary.

How old is Mary now?

38. You work in a diner waiting tables. One week, you earn twice as much on

Tuesday as you did on Monday, twice as much on Wednesday as you did on

Tuesday and twice as much on Thursday as you did on Wednesday. To celebrate

your improving earnings, you take Friday off and then calculate the average of

your earnings for the 5 days was $24/day. How much did you earn on

Wednesday?


103

Chapter 5: “Now I’m thinking of TWO numbers!”

When a single equation has two different algebraic variables, a door opens to what

seems like a new mathematical world. And yet, after we explore for a while, things

will start to look familiar. So let’s add another variable and take a look around.

5.1 Equations with Two Variables

Consider this equation: 2y – 4x = 10

Can we solve this equation? What would that mean? We would have to find values

of x and y that together make this equation a true mathematical sentence. I think

that if you play around for a while, you will find values of x and y that work. There

are lots of them. In fact, there are an infinite number of x and y pairs that fit this

equation.

Try to find your own pairs first. But if you are having trouble, here are three

examples that I came up with just by playing around with numbers.

For example… if 𝑦 = 5, then 2𝑦 = 10 by itself so we can let 𝑥 = 0 and we have a

solution: 𝑥 = 0, 𝑦 = 5

Or… if we let 𝑦 = 0 then we need to let 𝑥 = −2.5 and then that fits as well. So that

means that 𝑥 = −2.5, 𝑦 = 0 is another solution.

Or…if we let 𝑦 = 15, then 2𝑦 = 30 which is 20 more than we want. But 𝑥 = 5 will

then make 4𝑥 = 20 and since 30 – 20 = 10, these numbers work: 𝑥 = 5, 𝑦 = 15.

But there is another, more systematic way to go about finding solution pairs. You

could choose a value for one of the two variables and then substitute that value into

the equation. At that point, you would have an equation with just a single variable.

You know how to solve some of those! So let’s try that: let’s pick some x–values and

then solve for the y–values that go with them.

𝑆𝑎𝑦 𝑥 = 2. We chose this number arbitrarily.

2𝑦– 4𝑥 = 10 We were given this equation.

2𝑦– 8 = 10 We replaced the x with the value 2 and 4×2=8.

2𝑦 = 18 We added 8 to both sides.

𝑦 = 9 We divided both sides by 2.

So 𝑥 = 2, 𝑦 = 9 is one pair that satisfies the original equation.


We can repeat this same process as many times as we feel like: start with an x–

value, substitute it into the given equation and then solve for the y–value that goes

with it. (Hey! This is starting to sound familiar. We have y–values that “go with” x–

values…as if they had been matched…by some machine!)

KEY IDEAS IN THIS SECTION

Equations can have more than one variable.

If an equation has two variables, a solution consists of a pair of values – one for each

variable.

There may be LOTS of pairs of solutions.

Sometimes, you can find solution pairs by thinking and playing with numbers.

Sometimes, you can find solution pairs by choosing a value for one of the variables

and then solving for the other.

EXERCISES

1. For each given equation, check whether the proposed solution-pair satisfies the

equation.

a. 5𝑥 + 2𝑦 = 20 → 𝑥 = 2, 𝑦 = 5 ? b. 𝑚 + 3𝑛 = 0 → 𝑚 = 3, 𝑛 = 1 ?

c. 𝑥2 + 5𝑥 = 10𝑦 → 𝑥 = 2, 𝑦 = 4 ? d. 𝑝2 + 5𝑝 = 10𝑞 → 𝑝 = 5, 𝑞 = 5 ?

e. 2𝑥 = 𝑦2 + 4 → 𝑥 = 10, 𝑦 = 4 ? f. √𝑥 + 10 = 𝑦 − 5 → 𝑥 = 26, 𝑦 = 1 ?

2. For each these equations, play around with numbers until you find 3 different

solution pairs.

a. 4𝑥 − 3𝑦 = 12 b. 𝑦 + 4 = 5𝑥2 c. 𝑥𝑦 = 20

d. 𝑎

𝑏= 3 e. 𝐼 =

100

𝑟2 f. 𝑦 = 2𝑥 − 7


141

UNIT 3: Relationship Advice

There’s more to algebra than just solving

equations. Algebra is the language we use to

express mathematical relationships. In these

next few chapters, we will look at the

relationships that are most frequently

encountered, particularly in high school science

classes. This will give you a chance to see how we

apply the tools we have been developing in this

book.

For each relationship, we are going to look at a sample of (mostly simulated)

experimental results. We will see how the sample demonstrates that particular

mathematical relationship. Then, once we learn how to model that relationship

with an equation, we can apply our algebraic methods to solve new puzzles.

Many of my examples will refer to physics or chemistry that you may not yet have

learned. Please do not worry or be intimidated. One of the powerful things about

math is that it enables you to understand more than you think you can!


Chapter 6: Proportions

6.1 Direct Proportions and Classic Ratio Problems

Here are some things that math and science teachers have been known to say:

"The length of a building's shadow is directly proportional to its height."

"The volume of an ideal gas at constant pressure is directly proportional to the Kelvin temperature."

"The distance a car travels at a constant speed is directly proportional to the time of travel."

I know that I say these kinds of things regularly! In each of those sentences, you see

the words “directly proportional.” These are powerful code words, meant to trigger a

connection to a set of mathematical ideas. Teachers say “directly proportional,”

hoping that it will make you think about ratios, equations and graphs.

As beginners, my physics students usually tell me that “directly proportional”

means that when one thing goes up, so does the other. While that is true as far as it

goes, that property alone does not make things directly proportional. For example,

for much of your childhood, as your age increased, your height also increased. But

that alone does not make age and height directly proportional.

When we say that two things are directly proportional, it does not just mean that

when one increases, so does the other. It means that they both increase or decrease

by the same factor. If one triples, the other triples too. If one is divided by 5, so is

the other. So for your height to be proportional to your age, you would have to be

twice as tall at age 8 as you were at age 4. For some of you, that may in fact have

been true. But will your height double again between ages 8 and 16? Not likely.

There is a more concise way to say this: If two quantities are directly proportional, then as the quantities change, their ratio remains constant. Let’s examine that

definition and also review the topic of ratios.

The Ups and Downs of Inverse Variation 143

RATIOS REVISITED

When you want to know how much more one quantity is than another, you compare

them by subtracting. For example, if John is 15 years old and Mary is 10 years old,

then John is 15 – 10 = 5 years older. And, by the way, if subtracting results in a

negative number, that tells you that the first quantity was smaller than the second.

However, sometimes we are more interested in what we get when we compare the

two quantities using division. That kind of comparison is called a “ratio.” So in the

last example, the ratio of John’s age to Mary’s age is 15/10 = 1.5. We also write the

ratio in the form 15:10, using the colon to indicate that this is a ratio. Or we write

“15 to 10.” But when we evaluate that ratio, we use division. In this example, the

result of our division tells us that John is 1.5 times as old as Mary. And if John had

been younger than Mary, then the value of the ratio would have come out to be a

fraction less than one.

Subtraction tells us how much more, but a ratio tells us how many times more.

Let’s do an exercise that will help you to connect the two ways we have defined

“directly proportional” so far. It would defeat the purpose if I did this for you. So

follow the instructions below to see for yourself what happens.

A Quick Numeric Experiment

Pick two numbers, A and B. Calculate their ratio, A ÷ B.

Now we are going to increase A and B by the same factor. So choose a number to be

that factor. Multiply A and B each by that factor.

Let’s call the results “new A” and “new B.”

Calculate their ratio, “new A” ÷ “new B.” Aha!

As you have just seen, if two quantities change by the same factor, then their ratio

stays the same. Still not convinced? Pick new numbers and try again…


To explore this idea further, we are going to look at a classic example of direct

proportion: heights and shadows. Suppose you have been assigned to go outside on

a sunny afternoon and find an assortment of objects whose heights and shadows you

carefully measure. Here is an idealized version* of what your data might look like:

When you look at the data, it is not

surprising that the taller objects have

longer shadows. But are the heights

directly proportional to the shadows?

One way to check is to examine pairs of

data to see if they increased by the

same factor. For example, we could

compare the second row and the sixth

row. You can see that the height of the

object has tripled. And the same is

true for the lengths of the shadows.

Now let’s examine the ratios. If we use the second row to calculate the ratio of

shadow to height, we get 8

20=

2

5= .4. If we use the sixth row, we get

24

60 =

2

5 = .4

once again. Based on these two rows, we suspect that the two quantities are directly

proportional.

But what about the other rows? Did we happen to pick the only two pairs that

work? There is a way to check the data set as a whole. For each row, we calculate

the shadow-to-height ratio. Here is the chart again, this time with the calculated

ratios included:

* In real life, there would be measurement errors. These are unavoidable. And if you were

not quick enough taking measurements, we’d even have to account for the changing of the

angle of the setting sun! For now, I am shamelessly pretending that your measurements were

perfect so that the errors don’t distract us from the concept at hand. But you are welcome to

go outside and do this experiment yourself!

OBJECT

HEIGHT

(inches)

SHADOW

LENGTH

(inches)

1 Flower 15 6

2 Little Dog 20 8

3 Big Dog 35 14

4 Fence Post 40 16

5 Mail Box 50 20

6 Student 60 24

7 Taller Student 70 28

8 Lamp Post 110 44


As you can see, the heights got taller and the shadows got longer but the ratio

stayed constant! Thus, we can conclude that the height of an object is directly

proportional to the length of its shadow.

But why would we care? Well, once you know that two quantities are directly

proportional, you can use that information and a little bit of algebra to solve

interesting puzzles. For example, suppose your school was next to an old church

tower and you wanted to know how tall the tower was. It’s easy to measure the

length of its shadow, and then you are on your way.

A CLASSIC RATIO EXAMPLE

Jane is 67 inches tall. She measures her shadow and finds that it is 20 inches long.

Then, she measures the shadow of the church tower and finds that it is 190 inches

long. How tall is the tower?

We know that heights and shadow lengths are directly proportional. And there are

two objects we can compare: Jane and the tower. For one of the objects (Jane), we

are told both the height and the shadow length. For the other object (the tower), we

are told the shadow length and we are asked to find the height. But “directly

proportional” means that the ratio is the same! So the ratio we calculate based on

Jane’s height and shadow length must be equal to the ratio we would calculate

based on the tower’s height and shadow length.

We can also write that this way: 𝐻𝑒𝑖𝑔ℎ𝑡1

𝑆ℎ𝑎𝑑𝑜𝑤1=

𝐻𝑒𝑖𝑔ℎ𝑡2

𝑆ℎ𝑎𝑑𝑜𝑤2

which becomes: 67

20=

𝑥

190

EXERCISES

OBJECT

HEIGHT

(inches)

SHADOW

LENGTH

(inches)

Shadow-

to-height

ratio

1 Flower 15 6 6/15=.4

2 Little Dog 20 8 8/20=.4

3 Big Dog 35 14 14/35=.4

4 Fence Post 40 16 16/40=.4

5 Mail Box 50 20 20/50=.4

6 Student 60 24 24/60=.4

7 Taller Student 70 28 28/70=.4

8 Lamp Post 110 44 44/110=.4



Questions 15 – 20 present a pair of variables and two values for each variable. Use

that information to determine if the two variables could have an inverse-square

relationship.

15. When W = 30, Z = 10. And when W = 60, Z = 2.5.

16. When C = 100, D = 5. And when C = 10, D = 25.

17. You apply a 50-pound force on a rope which is connected to a pulley, pulling 12

feet of rope. As a result, the pulley lifts a 300-pound crate to a height of 2 feet.

18. Two oppositely charged objects attract each other with a force of 36 newtons

when they are separated by a distance of 2 meters. When the distance is

increased to 6 meters, the force decreases to 4 newtons.

19. When you are 10 meters away from a speaker at a rock concert, the sound

intensity is at the threshold of permanent damage. When you move to a

distance of 20 meters, the intensity drops to 25% of that threshold. (You may be

surprised to find that the level in decibels drops by a smaller amount, from 120

dB to 114 dB. This is because the decibel scale is an example of what is called a

“logarithmic scale,” an idea we will explore further in the exercises in Section

8.6.)

20. An object dropped from a point 1 Earth-radius away from the center of the

Earth accelerates downward at approximately 10 m/s2. An object dropped from a

point 2 Earth-radii away from the center of the Earth accelerates downward at

approximately 2.5 m/s2.

21. As in #19 above, when you are 10 meters away from a speaker at a rock concert,

the sound intensity is at the threshold of permanent damage, a level we can

designate as 100% intensity. If intensity varies inversely with the square of the

distance…

a. What will the intensity be at a point 15 meters away from the speaker?

b. At what distance would the sound have 10% of that maximum intensity?

22. A vertical pipe has a length of 12 meters and a radius of 5 centimeters. It is

filled with water.


a. If all of that water is poured into an empty barrel with a radius of 30

centimeters, to what depth will the water fill the barrel?

b. If instead, the water is poured into a different empty barrel which it fills to a

height of 1 meter, what is the radius of that barrel?

23. The three graphs below have a similar basic shape. One of them shows an

inverse proportion, one shows and inverse-square and one shows a relationship

you have not yet learned. Identify which is which.

A.

B.

C.

0

5

10

15

20

25

0 1 2 3 4 5 6 7 8 9 10

0

5

10

15

20

25

0 1 2 3 4 5 6 7 8 9 10

0

5

10

15

20

25

0 1 2 3 4 5 6 7 8 9 10


Now that you have learned to work with a variety of mathematical relationships, it’s

time to practice checking answers for “reasonableness.” Each of the following

problems comes with a proposed answer. Your goal is to determine if that answer is

reasonable, without actually solving the problem yourself. To do this:

A. Identify the relationship between the variables. The relationship and the

variables will have appeared in one of our earlier examples.

B. You will be given the initial and final values of one of the two variables. Note the

factor by which it has changed. Use that information to determine how the other

variable should change. Then see if the proposed solution agrees with that

prediction.

C. For the cases when the proposed solution is incorrect, give yourself extra credit if

you can identify the likely mistake that led to that incorrect answer.

24. Two flag poles stand next to each other. One is 10 feet tall and has a 4-foot long

shadow. The second pole is 14 feet tall. How long is its shadow?

Is it 8 feet?

25. A sample of gas has a volume of 3.6 liters and a pressure of 1.0 atmosphere. To

what volume must it be compressed to increase the pressure to 4.0 atmospheres?

Is it .9 liters?

26. A cylinder of clay has a radius of 3 inches and a height of 6 inches. It is

reshaped into a cylinder with a radius of 6 inches. What is the new height?

Is it 3 inches?

27. Four painters need five days to paint a house. If we hire six painters instead,

how long will it take them to paint the house?

Is it 7.5 days?

28. When you are 6 meters from a speaker, the music is

at an intensity that is 80% of the threshold for pain.

Showing good sense, you decide to move to a distance

where the intensity will be only 20%. How far away

from the speaker will you be then?

Is it 12 meters?


Chapter 8: Exponential Functions and Their Inverses

Exponential growth and its close relative, exponential decay, are found in an

incredibly wide variety of settings in math, science, and economics. From

population growth to radioactive decay to determining the future value of current

investments, many topics require an understanding of this new function.

8.1 The Power of Two

Suppose you had $1 in your pocket and were feeling pretty lucky. You might walk

into a casino and wager that $1 on the roulette wheel. Say you bet “red” and win,

bringing your wealth up to $2. So you go back the next day and successfully wager

all of your wealth again, bringing your total to $4. And you keep going back, once a

day, for twenty days. And every one of those days, you win, doubling your money.*

After 20 days, you would have a tidy sum.

DAY

You Started

This Day With: Today's Winnings: You Finished This

Day With: Total Accumulated

Winnings:

1 $1 $1 $2 $1

2 $2 $2 $4 $3

3 $4 $4 $8 $7

4 $8 $8 $16 $15

5 $16 $16 $32 $31

6 $32 $32 $64 $63

7 $64 $64 $128 $127

8 $128 $128 $256 $255

9 $256 $256 $512 $511

10 $512 $512 $1,024 $1,023

11 $1,024 $1,024 $2,048 $2,047

12 $2,048 $2,048 $4,096 $4,095

13 $4,096 $4,096 $8,192 $8,191

14 $8,192 $8,192 $16,384 $16,383

15 $16,384 $16,384 $32,768 $32,767

16 $32,768 $32,768 $65,536 $65,535

17 $65,536 $65,536 $131,072 $131,071

18 $131,072 $131,072 $262,144 $262,143

19 $262,144 $262,144 $524,288 $524,287

20 $524,288 $524,288 $1,048,576 $1,048,575

* This is my version of a very old, often told story. The classic version is about grains of rice

on a chessboard. You can find many descriptions of it on the web. And, in case my dubious

scheme actually tempts you, you should know that the series of successful wagers I describe

here is extremely unlikely even if you had a 50% chance on each wager – which you don’t.

Exponential Functions and their Inverses 185

As you can see, your wealth grew slowly at first but dramatically faster as the days

went by. In fact, each single day’s winnings are one dollar more than ALL of the

winnings up to that day. Look at Day 10 as an example. You won $512 that day

which is one more than the $511 you had won all together on the previous 9 days!

Now let’s just focus on two columns of the chart: the date, which

we will think of as the INPUT to a function, and your wealth at

the end of that day which we will call the OUTPUT. So the input

is the number of times you doubled your money and the output

is the total amount of money you now have in your pocket. Let’s

just look at the first 10 days.

For reasons that you will soon see, this kind of function is called

“exponential.” We can also represent it with a graph. For now,

when I draw the graph, I am not going to connect the dots. After

all, you only bet once each day. So the whole numbers are the

entire domain of this function. (But more on this later…)

Here is the graph:

The graph represents the same information as the chart, but it makes it easier to

observe the trend. You can see that your wealth grew slowly at first and then faster

and faster as the graph turns more steeply upward. Now that we have a graph and

a chart, let’s look for a rule.

$0

$200

$400

$600

$800

$1,000

$1,200

0 1 2 3 4 5 6 7 8 9 10

Wealth vs. Number of Days

DAY

You

finished

the day

with:

1 $2

2 $4

3 $8

4 $16

5 $32

6 $64

7 $128

8 $256

9 $512

10 $1,024


First, we should name the function. Let’s call it ‘W’ for wealth. And let’s call the

input ‘t’ since it represents a period of time, the number of days on which you have

doubled your money. So we are looking for an expression that will give us W(t).

We can express the rule for W(t) in words:

The output W(t) is the product of the number 2 multiplied by itself ‘t’ times.

For example, W(4) = 2×2×2×2 = 16. This can also be written as 24 = 16.

And W(7) = 2×2×2×2×2×2×2 = 128. This can be written as 27 = 128.

These last two examples suggest a more concise way to express the rule for W. We

can write:

W(t) is the value you get when you raise the base 2 to the exponent t.

Or more concisely: 𝑾(𝒕) = 𝟐𝒕.

Because the input to the function is used as the exponent, this kind of function is

called “exponential.” In this case, the base was 2, but exponential functions come in

lots of varieties with different bases. For example, in a different casino, you may

have made riskier bets with higher payoffs. If you tripled your money every day, the

wealth function would have been 𝑾(𝒕) = 𝟑𝒕. The graph of that function would look a

lot like the graph we already have.

Exponential functions start with that characteristic low slope and then turn steadily

steeper, assuming the base is greater than 1. Greater exponents cause faster

growth but as long as that exponent is even a little more than 1, the graph will have

that same characteristic shape. You’ll see this in the next example.



The next exercises are designed to lead to a discovery about the logarithm of a

product of two numbers.

26. a. Evaluate log 100 = ? b. Evaluate log 1000 = ?

c. The product 100 × 1000 =100,000. The log of that product, log 100,000 = ?

27. a. Evaluate log 10,000 = ? b. Evaluate log 0.1 = ?

c. The product 10,000 × 0.1 =1000. The log of that product, log 1000 = ?

28. You will need a calculator for this one.

a. Evaluate log 60 =? b. Evaluate log 25 =?

c. The product 60 × 25 = 1500. The log of that product, log 1500 = ?

29. Based on the results of the last 3 questions, propose a rule for the logarithm of a

product.

30. Design a series of calculations that will lead you to discover the quotient rule for

logarithms. Use exercises 26 – 28 as a template to get you started.

The rule that (I hope) you came up with for #29 is called the product rule for

logarithms. You will hear more about it in later math courses. It is this rule that

explains the operation of a slide rule, a type of mechanical multiplying machine. I

am not quite old enough to have used one of these in school. There was a time when

every math and physics student knew how to use one of these.

Now, it is nearly a forgotten art. But there are a number of websites with virtual

slide rules you can play with and tutorials on how to use them. The picture above is

from Derek Ross’s Virtual Slide Rule Gallery at:

http://www.antiquark.com/sliderule/sim/virtual-slide-rule.html.

http://www.antiquark.com/sliderule/sim/virtual-slide-rule.html

Exponential Functions and their Inverses 213

8.6 The Ballad of the Uncommon Logger

We started our exploration of exponential functions by

examining a function that helped you double your money.

As a function, the input is the exponent and the output is the

value when the base 2 is raised to that exponent.

But as you have seen in the last section, exponential functions

have inverse functions. We reverse the chart…

…and now the input is the value of the exponential

expression. And the output is the exponent you need when

the base is 2. This is still called a logarithm function, but it

is not the common logarithm. This one is called the “log base

2 of x.” It’s a new base, but the ideas remain the same.

When we were using 10 as the base, we said:

The base 10 logarithm assigns as its outputs the exponent you would use with a

base of 10 in order to get the input as the value of the exponential expression.

If 𝒚 = 𝐥𝐨𝐠𝟏𝟎 𝒙, then 𝟏𝟎𝒚 = 𝒙.

Now that we are using 2 as the base, we say:

The base 2 logarithm assigns as its outputs the

exponent you would use with a base of 2 in order to

get the input as the value of the exponential

expression.

If 𝒚 = 𝐥𝐨𝐠𝟐 𝒙, then 𝟐𝒚 = 𝒙.

The sentence to memorize this time is:

“Log2 8 = 3 because 23 = 8.”

𝑥 𝑦 = 2𝑥

1 2

2 4

3 8

4 16

5 32

6 64

7 128

𝑥 𝑦 = log2 𝑥

2 1

4 2

8 3

16 4

32 5

64 6

128 7

I see you are a logger…


EXAMPLES AND A NEAT CALCULATOR TRICK

Determine each of the following:

A. 𝐥𝐨𝐠𝟐 𝟏𝟔 B. 𝐥𝐨𝐠𝟐( 𝟏

𝟑𝟐) C. 𝐥𝐨𝐠𝟐 𝟐𝟓

If you happen to be familiar with the powers of 2 (and c’mon – who isn’t?) then the

first two can be done without a calculator.

A. We recognize that 24 = 16, so the logarithm is the value of the exponent, 4.

B. We recognize that 25 = 32, and we know the “one over” rule for negative

exponents, so the exponent (or logarithm) this time is -5.

But we will need some help with the third one. The exponent has to fall between 4

and 5 because 25 is more than 16 and less than 32. We can get that help from our

calculator, but it takes an extra step, using a trick that I call the Triumph

Algorithm after my high school classmate, James Triumph*.

The “log” button on your calculator always gives the common, base 10 logarithm.

There is another button labeled “LN” or “natural log.” That’s not what we want here

either. (The “natural log” is an important and interesting function, but it falls

beyond the scope of this book.) What we need is a log2 button and there isn’t one.

But there is a way around that. You can find the log base 2 of any number by

finding the common log and then dividing that answer by the common log of 2.

In equation form, it’s: 𝐥𝐨𝐠𝟐 𝒙 =𝐥𝐨𝐠𝟏𝟎 𝒙

𝐥𝐨𝐠𝟏𝟎 𝟐.

We will explore why this works soon. But first let’s verify that it does work. To

start, let’s use this algorithm to check our first answer above.

We wanted log2 16. Log10 16 = 1.2041 and log10 2 = .3010.

Log10 16

log10 2 =

1.2041

.3010= 4.00 which is what we got earlier just by knowing our powers of 2.

* I’m not saying he invented it. But I know that James figured it out for himself back in 10th

grade and then he taught it to me. An “algorithm” is a series of steps which we use to solve a

problem. So I guess you can think of this one as a logarithm algorithm.



0

5

10

15

20

25

30

0 2 4 6 8 10

0

2

4

6

8

10

0 2 4 6 8 10

0

5

10

15

20

25

0 1 2 3 4 5

Chapter 3

1.

𝐻(𝑥) = 2𝑥 + 5

2. Answers will vary depending on chosen values. Here is one possible answer:

3. Changing the letter you use to represent the input does not change the function at all.

4. Changing the letter you use to name the function changes…the name of the function! But

that’s all it changes.

5. 𝑃(𝐿) = 4𝐿.

x H(x)

0 5

2 9

4 13

6 17

8 21

10 25

x k(x)

2 8

4 6

6 4

8 2

10 0

L P(L)

1 4

2 8

3 12

4 16

5 20

Answers and Hints 231

0

100

200

300

400

0 5 10 15 20

6.

7. a. The function is named “R.”

b. Its inputs are represented with the letter q.

c. To find the output, R triples the inputs and then subtracts 1.

8. Tom doubles the inputs and then subtracts 2, or T(x) = 2x – 2.

9. Jerry subtracts 1 and then squares, or J(x) = (x – 1)2.

10. a. 52 b. 100 c. 200 d. 15 e. 1024

11. a. 64 b. 76 c. 32 d. 7

12. a. (3 × 5 + 1)2 = 256 b. (3 × (5 + 1))2

= 324

c. 3 × (5 + 1)2 = 108 d. 3 × 52 + 1 = 76

13. For d, above, the sequence of operations in the description followed the hierarchy of the

order of operations so no parentheses were required.

14 a. 25+16

4= 12 b.

2(3+5)

4= 64

c. 23×(2+1) = 512

15. For a, above, the sequence of operations in the description followed the hierarchy of the

order of operations so no parentheses were required.

16. a. 125 b. 53 c. 215 d. 52

17. a. 35 b. 36 c. 8 d. 16

18. a. 3 × 4 + 4 × 4 b. 4 × 4 + 3 × 4

c. 4 + 3 × 4 + 4 d. 4 + 3 ÷ 4 × 4

t x(t)

0 100

5 150

10 200

15 250

20 300


19. a. 72 + 3 b. 3 + 72 c. 3 + 72 × 2 d. 23+ 7 e. ???

20. I don’t know how to rewrite 19e without parentheses. This new system would require a

new symbol for handling exponents.

21. a. 𝑎(𝑥) = 2𝑥 + 6 +𝑥

4 , 𝑎(6) = 2 × 6 + 6 +

6

4= 19.5

b. 𝑝(6) = 2 × 6 + 6 = 18, 𝑞(6) =6

4= 1.5, 𝑝(6) + 𝑞(6) = 18 + 1.5 = 19.5

22. a. 𝑏(𝑥) =𝑥

4

2𝑥+6 , 𝑏(12) =

12

4

2×12+6=

3

30=

1

10

b. 𝑞(12) = 12

4= 3, 𝑝(12) = 2 × 12 + 6 = 30,

𝑞(12)

𝑝(12)=

3

30=

1

10

23. a. 𝑐(𝑥) = 2 (𝑥

4) + 6 , 𝑐(20) = 2 (

20

4) + 6 = 16

b. 𝑞(20) =20

4= 5 , 𝑝(5) = 2 × 5 + 6 = 16

24. a. 𝑑(𝑥) =2𝑥+6

4 , 𝑑(5) =

2×5+6

4= 4

b. 𝑝(5) = 2 × 5 + 6 = 16 , 𝑞(16) =16

4= 4

25. 𝑔(𝑥) =(𝑥+4)3

2 26. ℎ(𝑥) =

𝑥3+4

2

27. 𝑗(𝑥) =(

𝑥+4

2)

2 or 𝑗(𝑥) =

𝑥+4

4 28. 𝑘(𝑥) = (𝑥 + 4)3 + 4

29 and 30…answers vary

31. 𝑓(10) =10−5

2= 2.5 , 𝑔(2.5) = 2 × 2.5 + 5 = 10

32. ℎ(𝑥) = 2 (𝑥−5

2) + 5 = 𝑥 − 5 + 5 = 𝑥

33. 𝑔(6) = 2 × 6 + 5 = 17 , ℎ(17) =17−5

2= 6

34. 𝑘(𝑥) =(2𝑥+5)−5

2=

2𝑥

2= 𝑥

35. 𝐿(𝑀(𝑥)) = 𝑥 36. 𝑀(𝐿(𝑥)) = 𝑥

37. These are the “natural solutions.” For “wise-guy” or “clever” solutions, see

advancedmathyoungstudents.com

a. ℎ(𝑥) = 𝑥2, 𝑔(𝑥) = 𝑥 + 2 b. ℎ(𝑥) = 𝑥 − 5, 𝑔(𝑥) =𝑥

4 c. ℎ(𝑥) = 𝑥 + 8, 𝑔(𝑥) = 𝑥3

Answers and Hints 233

Chapter 4

1. c, d, e, g and j are equations, a, b and f are expressions and h and i are open to

interpretation.

2. In each case, just substitute the proposed solution into the equation and see if the result is

true. You should find that a, c, d, e and f are all correct. Note that g has no solutions so any

value you try leads to false statement.

3. 12𝑥 + 48 4. 20𝑥 − 100 5. 31 − 3𝑥

6. 16𝑥 − 2 7. 2𝑥 − 2 8. 3𝑥2 + 7

9. 2𝑥 − 4 10. 2𝑥 + 4 11. 24 − 13𝑥

12. no change 13. 26 + 17𝑥 14. 100 − 4𝑥2

15. 7(𝑥 + 7) 16. 4(𝑥 − 5) 17. 3(2𝑎 + 9)

18. 3(4 − 𝑥) 19. no change 20. 2(4𝑥 + 𝑦)

For 21 – 26, a calculator reveals the answer. Here are some suggested approaches to the

mental math. There more than one way to do these, so if you have found another way than

the one I suggest, that’s good.

21. 33 × 9 = (30 + 3) × 9 = 30 × 9 + 3 × 9…

22. 33 × 11 = 33 × (10 + 1) = 33 × 10 + 33 × 1…

23. 48 × 22 = (50 − 2) × 22 = 50 × 22 − 2 × 22 = 50 × (20 + 2) − 44 …

24. 102 × 11 = (100 + 2) × 11 = 100 × 11 + 2 × 11 …

25. (30 + 4) × (20 + 6) = (30 + 4) × 20 + (30 + 4) × 6 = 30 × 20 + 4 × 20 + 30 × 6 + 4 × 6…

26. (50 + 5)(50 − 5) = (50 + 5) × 50 − (50 + 5) × 5 = 50 × 50 + 50 × 5 − 50 × 5 − 5 × 5 …


For 27 – 38, here are suggested “translations” into algebra. Again, there is more than one

way. Then, on the next page, you will find proposed equation ladders for each of these.

27. Let x be the original number. 3𝑥 − 10 = 𝑥

28. Let m be the original number of marbles. 1

2𝑚 − 20 =

1

4(𝑚 − 20)

29. Let C be the number of cars. 4𝐶 × 5 + 2(24 − 𝐶) × 5 = 380

30. Let n be the first of the three even integers. 𝑛 + (𝑛 + 2) + (𝑛 + 4) = 612

31. Let p be the LARGEST of the 4. 𝑝 + (𝑝 − 2) + (𝑝 − 4) + (𝑝 − 6) = 𝑝 + 45

32. Let S be the original price of the sweater. 1.2𝑆 + 10 = 2𝑆

33. Let F be the number of freshman. 𝐹 + 2𝐹 + 4𝐹 + 12 = 68

34. Let T be the number of texts. 5 + .02𝑇 = 8 + .01𝑇

35. Let D be the length of the trip in miles. 𝐷

60=

𝐷

40− .25

36. See Solution guide at www.advancedmathyoungstudents.com

37. Let M be Mary’s age. 3𝑀 + 10 = 2(𝑀 + 10)

38. Let x be what you earned on Monday. 𝑥+2𝑥+4𝑥+8𝑥+0

5= 24

Equations ladders are on the next page. I did not always look for the shortest ladder. I tried

to choose steps that were easy to follow.


http://www.advancedmathyoungstudents.com/

257

INDEX

A

absolute value, 99

algebraic expression

definition of, 52

assignments and not options rule, 12

B

Boyle’s Law, 170

C

Cantor, Georg, 206

Charles’s Law, 154

combining like terms, 82

connecting dots, 31

D

direct proportion, 6, 142, 144, 148, 150,

151, 158

direct-square variation, 6, 163, 164

Dirichlet Function, 206

distributing the minus sign, 81

distributive property

and mental math, 79

domain

definition of, 10

example of, 19

E

equation

algebraic, 6, 68, 72, 76

definition, 68

linear, 6

quadratic, 99

two-variable, 103

equation ladder, 74, 76

exponential decay, 187

exponential expression, 191, 211

exponents

definition, 191

fractional, 200, 202

laws of, 191, 199

negative, 196, 199

zero as, 195, 199

F

functions

combining, 23

composition of, 6, 24, 25

exponential, 7, 183, 185, 188

inverse, 6, 13, 135, 137, 139, 207, 217

linear, 109, 112, 154

nested, 24

notation of, 21

one-to-one, 13

rules of, 47

G

graph of a function

constructing, 29

using slope, 119

graphs

representing functions with, 6

representing inverse functions with, 40

representing logarithmic functions

with, 210

H

Hippasus, 205


horizontal line test, 44, 46, 229

hyperbola, 171

I

intersection

of two lines, 123, 127

inverse proportion. See inverse variation

inverse square variation, 6, 176

inverse variation, 6, 169, 173

irrational numbers, 205

isolating the variable, 106

K

Kepler’s Third Law, 221

L

line of symmetry, 43

logarithm, 7, 207, 211, 217

common, 208, 217

M

match-making

using functions for, 10

O

order of operations, 52

output rule

for combined functions, 59

for composition of functions, 60

for functions, 47

P

parabola, 165, 167

parentheses

implied, 55

PEMDAS, 52, 79

per cents, 186

proportionality

constant of, 151, 158, 165

Pythagorean Theorem, 101

Q

quadratic formula, 7

R

range

definition of, 10

example of, 19

ratio, 143, 145, 147, 148

restricting the domain, 46, 95

rise over run, 114, 120

S

SAT, 5, 71

simultaneous equations

system of, 124

simultaneous solution, 124

slope, 113, 120

negative, 117, 120

slope formula, 118

slope-intercept, 120

solution

definition, 73

solutions

verifying, 74

solving for y, 106

square roots, 96, 100

squaring, 94, 100

substitution, 124, 125, 127, 130, 132

T

The New Math SAT Game Plan, 5

transformations, 92

of equations, 78, 86

Triumph Algorithm, 214

Triumph, James, 214

259

V

vertical line test, 34, 44, 45, 227, 228

W

word ladder puzzles, 73

word problems, 6, 71, 74, 101, 129, 130,

132, 133

Y

y = x line, 43, 95

y-intercept, 35, 113, 120


And finally…

I certainly hope that your evolving mastery of the language of algebra will enable

you to sail through the SATs with ease. But when the time comes, you may also

want to pick up a few non-algebraic strategies as well.

Available at Amazon.com